Visualize Fundamental Homomorphism Theorem for $\phi: A_4 \rightarrow C_3$ 
Question 1. How do you see $\ker\phi = V_4 $ = Klein 4 group ? Book doesn't give formula for $\phi$?
 Question 2. What's $H$ in $i(aH) = \phi(a)$? I think $H = \ker\phi$ ?
 Question 3. Why is $i: \frac{A_4}{\ker\phi} \to C_3$ defined as $i(aH) = \phi(a)$ ? Why not just $i(aH) = a$ ?  
This is from Nathan Carter page 169  Visual Group Theory.
Question 4. page 167 says 
When $\phi$ is not an embedding, somewhere it must collapse two domain elements to
one codomain element. In fact, because quotient maps follow a repeating pattern, every
coset of $\ker\phi$ will have at least two elements in it.
Can someone please explain this last sentence? 
 A: Often times we find ourselves asking whether a homomorphism between two groups exist. There is a method which lends itself to be more intuitive. And this method is exemplified by an example. Imagine a square (anchored in the usual way in $S^1$ if you like). Two groups that naturally act on the square are $C_4$ and $D_{2\cdot 4}$ (order 8). Now imagine the two diagonals of the square. Each time you act on the square, you also act on these diagonals. The automorphism group of these diagonals are $C_2$. Thus there is a homomorphism from $C_4$ to $C_2$ and a homomorphism from $D_{2\cdot 4}$ to $C_2$. And you could build these homomorphisms on the generators watching what they do to the diagonals. 
This method generalizes. To determine whether there is a homomorphism between two groups, you should try to find a 'structure' that one group naturally acts on, and another structure that can be built out of the first structure which the second group naturally acts on. In our example the first structure was the square and the second the diagonals.
Now we can apply this principle to $S_4$ and $S_3$. $S_4$ is typically seen as the permutations of the set $T=\{1,2,3,4\}$---this is our first structure. Now we build our second structure. There are three ways to partition $T$ into pairs of pairs. We list them: $a_1=\{1,2\}\cup\{3,4\}$, $a_2=\{1,3\}\cup\{2,4\}$, and $a_3=\{1,4\}\cup\{2,3\}$. Now imagine the set $H$ whose elements are $a_1, a_2$ and $a_3$. $S_3$ naturally acts on $H$. And we can see that every action of $S_4$ on $T$ induces an action on $H$. For example let's take the element $(1\;2)\in S_4$. Then you can see that $(1\; 2)$ induces a map which sends $a_1$ to $a_1$, $a_2$ to $a_3$, and $a_3$ to $a_2$. This map corresponds to a map in $S_3$. And this is how the homomorphism is built. Can you see the kernel of this map? It is $V_4$: the elements $(1\;2)(3\;4)$, $(1\;3)(2\;4)$, $e$, and $(1\;4)(2\;3)$.
EDIT 
Your map is constructed differently. $A_4$ can be realized as the rigid motions of the tetrahedron (thus not including reflection). There is an octahedron inside the tetrahedron which is preserved by every rigid motion of the tetrahedron. And there are three diagonals inside the octahedron which are preserved by every rigid motion of the octahedron. Thus every rigid motion of the tetrahedron permutes these three line segments. And this gives rise to your homomorphism. You can realize these line segments another way. Every edge of the tetrahedron opposes another edge which is affinely perpendicular to it. Draw the line segment from each of these edges midpoints. 

A: The standard proof of Lagrange's Theorem shows that all cosets of a given subgroup have the same cardinality (size).  In the quotient homomorphism $\phi$ described in your illustration, the subgroup $K = \ker \phi$ has $|K| > 1$, so every coset will too.
Here's the argument, briefly.  If $g_1K$ and $g_2K$ are cosets of $K < G$, then the map $x \mapsto g_2g_1^{-1}x$ is a bijection $g_1K \overset{\sim}{\to} g_2K$.
